37.8 ~g ~N₂ O₅$37.8 \mathrm{~g} \mathrm{~N}_{2} \mathrm{O}_{5}$ was taken in a 1 ~L$1 \mathrm{~L}$ reaction vessel and allowed to undergo the following reaction at 500 ~K$500 \mathrm{~K}$: 2 N _ 2 O _ 5 (g) leftharpoons 2 N _ 2 O _ 4 (g) + O _ 2 (g)$$2 \mathrm {N} _ {2} \mathrm {O} _ {5 (\mathrm {g})} \rightleftharpoons 2 \mathrm {N} _ {2} \mathrm {O} _ {4 (\mathrm {g})} + \mathrm {O} _ {2 (\mathrm {g})}$$ The total pressure at equilibrium was found to be 18.65 bar. Then, Kp = _ _ _ _ × 10⁻²$\mathrm{Kp} = \_ \_ \_ \_ \times 10^{-2}$ [nearest integer] Assume N₂O₅$\mathrm{N}_2\mathrm{O}_5$ to behave ideally under these conditions Given: R = 0.082$\mathrm{R} = 0.082$ bar L mol⁻¹ K⁻¹$\mathrm{L} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$

Related Formula

Core Logic

For the given gaseous equilibrium reaction:

First, find the change in the number of moles of gas (Δ ng$\Delta n_g$):

Step 1: Calculation

Substitute the given values into the Kₚ - Kc$K_p - K_c$ relationship: Kₚ = 0.492$K_p = 0.492$ R = 0.082$R = 0.082$ T = 300 K$T = 300\text{ K}$

Kc = 0.02$K_c = 0.02$

Converting to the requested format (x × 10⁻²$x \times 10^{-2}$):

So, the value is 2.

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Henderson-Hasselbalch Equation for Acidic Buffers:

Core Logic

Let the volume of 1M Benzoic acid be Vₐ$V_a$ mL and the volume of 1M Sodium benzoate be Vₛ$V_s$ mL. Total volume = Vₛ + Vₐ = 300 mL$V_s + V_a = 300\,\text{mL}$.

Millimoles of acid = 1 × Vₐ = Vₐ$1 \times V_a = V_a$ Millimoles of salt = 1 × Vₛ = Vₛ$1 \times V_s = V_s$

Applying Henderson's Equation:

Step 1: Calculate Volume Ratio

Since 2 ≈ 0.3$\log 2 \approx 0.3$, we have:

Vₛ = 2 Vₐ$V_s = 2 V_a$

Step 2: Substitute and Solve

We know Vₛ + Vₐ = 300$V_s + V_a = 300$ Substituting Vₛ = 2 Vₐ$V_s = 2 V_a$:

3 Vₐ = 300$3 V_a = 300$

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

pH + pOH = 14$pH + pOH = 14$

Core Logic

Precipitation begins just when the ionic product equals the solubility product (Qₛₚ = Kₛₚ$Q_{sp} = K_{sp}$).

Step 1: Calculating required [OH-]

Given [Mg²⁺] = 0.10 M$[Mg^{2+}] = 0.10 \text{ M}$

Step 2: Finding pH

pH = 14 - pOH$pH = 14 - pOH$ pH = 14 - 5 = 9$pH = 14 - 5 = 9$

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

where Pᵢ$P_i$ is the partial pressure of component i$i$.

Step 1: Setting up the ICE Table

For the reaction A(g) leftharpoons B(g) + (1)/(2) C(g)$A_{(g)} \rightleftharpoons B_{(g)} + \frac{1}{2} C_{(g)}$

Let initial moles of A = 1$A = 1$. At equilibrium: Moles of A = 1 - α$A = 1 - \alpha$ Moles of B = α$B = \alpha$ Moles of C = (α)/(2)$C = \frac{\alpha}{2}$

Total moles at equilibrium = (1 - α) + α + (α)/(2) = 1 + (α)/(2) = (2 + α)/(2)$(1 - \alpha) + \alpha + \frac{\alpha}{2} = 1 + \frac{\alpha}{2} = \frac{2 + \alpha}{2}$

Step 2: Calculating Partial Pressures

Using mole fraction ×$\times$ Total Pressure (P$P$): PA = (1 - α)/(1 + (α)/(2)) · P$P_A = \frac{1 - \alpha}{1 + \frac{\alpha}{2}} \cdot P$ PB = (α)/(1 + (α)/(2)) · P$P_B = \frac{\alpha}{1 + \frac{\alpha}{2}} \cdot P$ PC = ((α)/(2))/(1 + (α)/(2)) · P$P_C = \frac{\frac{\alpha}{2}}{1 + \frac{\alpha}{2}} \cdot P$

Step 3: Calculating Kp

Since 1 + α/2 = (2+α)/(2)$1 + \alpha/2 = \frac{2+\alpha}{2}$, the √(2)$\sqrt{2}$ in denominator cancels out perfectly leaving:

Chapter Mix

Class 11 Chemistry: Equilibrium

Related Formula

Core Logic

Chapter Mix

Class 11 Chemistry: Equilibrium